Eigenspinors of Sy (Griffiths 4.29)

  • Thread starter Rubiss
  • Start date
In summary: So basically, the same thing happens as in part (b), where you just set \alpha = 1 to get \beta = \pm i.
  • #1
Rubiss
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Homework Statement



Griffiths Quantum Mechanics, 2nd edition:
Calculate the eigenvalues and eigenspinors of Sy

Homework Equations



See attached PDF. Everything is in the PDF.

The Attempt at a Solution



See attached PDF. All of my steps are written out. I promise.


Parts 1, 2, and 3 are all in the attached PDF, I promise. I wrote out all of the steps in my work, so you can follow along. I am wondering where I made a math error in calculating the eigenspinors of Sy. I have some questions about b and c, and my work is shown for those parts as well. Hopefully someone can help.

I'm really sorry I couldn't write all of my math in this message board. I tried at first, but I couldn't get the matrices to work. So I wrote everything in my latex editor, and the result is the attached PDF.

Thanks.
 

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  • #2
Did you notice your equation (4) and your solution for [itex]\alpha[/itex] at the bottom of page 1 are the same?

Eigenvectors are unique only up to a multiplicative constant, so just arbitrarily set one of the variables to 1 and solve for the other and then normalize.

Your results are correct, but you have them swapped. If you multiply by [itex]\pm i[/itex], you'll get the answers you're looking for.Your work for part (b) is fine.Your work for part (c) isn't. You don't find probabilities by calculating the expectation value. You expand the state in terms of the eigenstates of the observable and then pick off the coefficients.
 
  • #3
vela said:
Did you notice your equation (4) and your solution for [itex]\alpha[/itex] at the bottom of page 1 are the same?

Vela, thank you for responding. I greatly appreciate it. I didn't notice that equation (4) and my solution for [itex]\alpha[/itex] was the same. Thanks for pointing that out.

vela said:
Eigenvectors are unique only up to a multiplicative constant, so just arbitrarily set one of the variables to 1 and solve for the other and then normalize.


I not sure I understand what you are saying. Why can't I use my method of plugging alpha into my eigenspinor, and obtaining the correct result? I can use equation (4) to solve for [itex]\beta[/itex] and obtain [itex]\beta=\pm i \alpha[/itex]. When I plug in these two values of [itex]\beta[/itex] into my eigenspinor, I obtain the correct result. Why should these way be any different?

vela said:
Your results are correct, but you have them swapped. If you multiply by [itex]\pm i[/itex], you'll get the answers you're looking for.


I'm not sure I understand how my results are correct. Aren't the eigenspinors I obtained [equations (9) and (12)] fundamentally different from the correct eigenspinors [equations (13) and (14)]?

vela said:
Your work for part (b) is fine.

But is there any other way to solve this problem without using the identity operator?

vela said:
Your work for part (c) isn't. You don't find probabilities by calculating the expectation value. You expand the state in terms of the eigenstates of the observable and then pick off the coefficients.

Ok. From part b, I have expanded the state in terms of the eigenstates of the observable Sy. How do I expand the state in terms of the eigenstates of the observable Sy2?
 
  • #4
Rubiss said:
I not sure I understand what you are saying. Why can't I use my method of plugging alpha into my eigenspinor, and obtaining the correct result? I can use equation (4) to solve for [itex]\beta[/itex] and obtain [itex]\beta=\pm i \alpha[/itex]. When I plug in these two values of [itex]\beta[/itex] into my eigenspinor, I obtain the correct result. Why should these way be any different?
I was just pointing out that most of the work you did after getting equation (4) is unnecessary. Once you have (4), just set, say, [itex]\alpha = 1 [/itex] and you'll get [itex]\beta=\pm i[/itex].
I'm not sure I understand how my results are correct. Aren't the eigenspinors I obtained [equations (9) and (12)] fundamentally different from the correct eigenspinors [equations (13) and (14)]?
Take equation (9). Multiply it by -i. You get (14). Similarly, Multiply (12) by i, and you get (13).
But is there any other way to solve this problem without using the identity operator?
If the state of the system is [itex]| \psi \rangle [/itex], the probability of finding the system in the state [itex]| \phi \rangle[/itex] is [itex]|\langle \phi | \psi \rangle |^2[/itex]. This is true for any two states. When you use eigenstates for [itex]|\phi\rangle[/itex], you essentially get your results. Really, all you need are the coefficients in the expansion, so that's all you need to calculate.
Ok. From part b, I have expanded the state in terms of the eigenstates of the observable Sy. How do I expand the state in terms of the eigenstates of the observable Sy2?
What are the eigenstates of Sy2 and their associated eigenvalues?
 
  • #5
vela said:
What are the eigenstates of Sy2 and their associated eigenvalues?

I calculated that Sy2=[itex]\frac{\hbar^{2}}{4}[/itex](1 0; 0 1). (That matrix is the identity matrix). I found that there is only one eigenvalue for this matrix: [itex]\frac{\hbar^{2}}{4}[/itex]. Since this is the only eigenvalue of Sy2, this is the only value I could obtain if I were to measure Sy2. Is this correct? Have I done enough work, or do I need find the eigenstates of Sy2?
 
  • #6
Yup, that's right.
 

What are eigenspinors?

Eigenspinors are a type of mathematical object that are used in quantum mechanics to describe the behavior of spin particles. They are eigenfunctions of the spin operator, meaning that they represent the possible spin states that a particle can have.

How are eigenspinors related to Sy (Griffiths 4.29)?

In Griffiths 4.29, eigenspinors are used to describe the spin states of a particle in a magnetic field. The spin operator, Sy, is one of the components of the total spin operator and can be used to determine the eigenvalues and eigenspinors for a given spin state.

What is the significance of eigenspinors in quantum mechanics?

Eigenspinors are important in quantum mechanics because they represent the possible spin states of a particle, which is a fundamental property in quantum mechanics. They also have applications in calculating probabilities and predicting the behavior of particles in quantum systems.

How do you calculate the eigenspinors of a given spin state?

To calculate the eigenspinors of a given spin state, you first need to determine the eigenvalues of the spin operator for that state. Then, you can use these eigenvalues to solve the eigenspinor equation and determine the corresponding eigenspinors.

Can eigenspinors be used to describe particles other than spin particles?

Yes, eigenspinors can be used to describe other types of particles, such as fermions and bosons, as long as they have a spin property. They are also used in fields outside of quantum mechanics, such as in the study of crystal structures and their symmetries.

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